GATE 2023 EC Q.46: solution and answer

GATE 2023 · ECQ.46 · 2 marks · MCQ

The h-parameters of a two port network are shown below. The condition for the maximum small signal voltage gain $\frac{V _{o u t}}{V _{s}}$ is

GATE 2023 EC question 46: Input side (left): an independent voltage source V_s (a circle with no polarity marks) in the left vertical branch. — Figure from the GATE 2023 EC paper. It is part of the question.

(A)
$h_{11} = 0, h_{12} = 0, h_{21} = very high and h_{22} = 0$
(B)
$h_{11} = very high, h_{12} = 0, h_{21} = very high and h_{22} = 0$
(C)
$h_{11} = 0, h_{12} = very high, h_{21} = very high and h_{22} = 0$
(D)
$h_{11} = 0, h_{12} = 0, h_{21} = very high and h_{22} = very high$

Answer and solution

ANSWER

(A), for the standard h-parameter model the stem names (output source $h_{21} i_{1}$ ). There $A_{v} = \frac{- h _{21} R _{L}}{h _{11} ( 1 + h _{22} R _{L} ) - h _{12} h _{21} R _{L}}$ , and (A) makes it largest. As printed, the source is labelled $h_{21} v_{1}$ . With that label the gain does not depend on $h_{11}$ or $h_{12}$ , so (A), (B) and (C) tie, and the official key awards marks to all candidates.

Official answer: marks to all candidatesSource: Official answer key, GATE 2023 EC (IIT Kanpur)

What the stem fixes. The stem says the network is drawn with its h-parameters. By definition, its port equations are then $v_{1} = h_{11} i_{1} + h_{12} v_{2}, i_{2} = h_{21} i_{1} + h_{22} v_{2}$ Here $h_{21}$ is the ratio $i_{2} / i_{1}$ with the output shorted ( $v_{2} = 0$ ), which is a pure number. The input half of the figure is exactly the first equation: $h_{11}$ in series with the dependent source $h_{12} v_{2}$ .
As printed, the figure has no single answer. The output source is printed as $h_{21} v_{1}$ , not $h_{21} i_{1}$ . Take that label literally. Then $v_{1} = V_{s}$ , because its + and − marks sit across the source terminals. KCL at the output top node, with $v_{2} = - i_{2} R_{L}$ , gives $v_{2} = - R_{L} (h_{21} V_{s} + h_{22} v_{2})$ , so $\frac{V _{o u t}}{V _{s}} = \frac{- h _{21} R _{L}}{1 + h _{22} R _{L}}$ Neither $h_{11}$ nor $h_{12}$ appears. Options (A), (B) and (C) all have $h_{21}$ very high and $h_{22} = 0$ , so they give the same gain, $- h_{21} R_{L}$ (reading $h_{11} = 0$ as the limit $h_{11} \to 0$ ). Only (D) gives less. With the printed label, three options are equally correct, and the official key awards marks to all candidates. The label is also inconsistent in itself: a source $h_{21} v_{1}$ would make $h_{21}$ a conductance in siemens, which an h-parameter is not. The rest of this solution uses the h-parameter model the stem names, with the output source $h_{21} i_{1}$ .
Source and load. The ideal source drives port 1 directly, so $v_{1} = V_{s}$ . At port 2, $i_{2}$ is drawn flowing from $R_{L}$ into the network. The current down through $R_{L}$ is therefore $- i_{2}$ , and $v_{2} = V_{o u t} = - i_{2} R_{L}$
Output side. Substitute $i_{2} = - v_{2} / R_{L}$ into $i_{2} = h_{21} i_{1} + h_{22} v_{2}$ : $- \frac{v _{2}}{R _{L}} = h_{21} i_{1} + h_{22} v_{2} \Rightarrow i_{1} = - \frac{( 1 + h _{22} R _{L} ) v _{2}}{h _{21} R _{L}}$
Input side. Substitute this $i_{1}$ into $V_{s} = h_{11} i_{1} + h_{12} v_{2}$ : $V_{s} = v_{2} (h_{12} - \frac{h _{11} ( 1 + h _{22} R _{L} )}{h _{21} R _{L}})$ The voltage gain is then $A_{v} = \frac{V _{o u t}}{V _{s}} = \frac{- h _{21} R _{L}}{h _{11} ( 1 + h _{22} R _{L} ) - h _{12} h _{21} R _{L}}$
How to compare the options. All four options have $h_{21}$ very high. Each of (B), (C) and (D) differs from (A) in exactly one parameter, so compare them with (A) one parameter at a time.
$h_{22}$ : (A) against (D). With $h_{12} = 0$ , $∣ A_{v} ∣ = \frac{h _{21} R _{L}}{h _{11} ( 1 + h _{22} R _{L} )}$ A very high $h_{22}$ divides the gain by $1 + h_{22} R_{L}$ , because the shunt admittance carries away current that would otherwise flow through $R_{L}$ . So $h_{22} = 0$ , and (D) is out.
$h_{12}$ : (A) against (C). With $h_{22} = 0$ , $A_{v} = \frac{- h _{21} R _{L}}{h _{11} - h _{12} h _{21} R _{L}}$ As $h_{11} \to 0$ this tends to $\frac{1}{h _{12}}$ , which is tiny when $h_{12}$ is very high. The feedback source $h_{12} v_{2}$ caps the gain at about $1/ h_{12}$ , and only $h_{12} = 0$ removes the cap. So (C) is out.
$h_{11}$ : (A) against (B). With $h_{12} = 0$ and $h_{22} = 0$ , $A_{v} = - \frac{h _{21} R _{L}}{h _{11}}$ The input current is $i_{1} = V_{s} / h_{11}$ , and the output source feeds $h_{21}$ times that current into $R_{L}$ . For the same $h_{21}$ , a very high $h_{11}$ chokes $i_{1}$ and lowers the gain, while $h_{11} \to 0$ raises it without bound. So (B) is out.
Answer (A): $h_{11} = 0$ , $h_{12} = 0$ , $h_{21}$ very high and $h_{22} = 0$ , which gives $A_{v} = - h_{21} R_{L} / h_{11}$ . Read $h_{11} = 0$ as the ideal limit $h_{11} \to 0$ . Exactly zero, together with $h_{12} = 0$ , would put a short circuit across the ideal source $V_{s}$ .

KEY IDEA

Write the gain in h-parameters, $A_{v} = \frac{- h _{21} R _{L}}{h _{11} ( 1 + h _{22} R _{L} ) - h _{12} h _{21} R _{L}}$ . Each wrong option changes exactly one parameter from (A), so compare one parameter at a time. $h_{22}$ is a shunt that steals load current. $h_{12}$ is feedback that caps the gain near $1/ h_{12}$ . $h_{11}$ sets $i_{1} = V_{s} / h_{11}$ , the current that the output source multiplies by $h_{21}$ .

WATCH OUT FOR

Choosing (B) from the rule that a voltage amplifier needs a high input resistance. That rule is about not loading a source that has its own internal resistance. Here $V_{s}$ is ideal, so there is no loading to avoid: $h_{11}$ only sets $i_{1} = V_{s} / h_{11}$ , and a larger $h_{11}$ means less gain. A second trap is solving the figure literally with $h_{21} v_{1}$ , which makes $h_{11}$ and $h_{12}$ irrelevant, and then quietly switching to $h_{21} i_{1}$ to rule out an option.

From the GATE 2023 EC paper. Every question of this paper · Download the full paper · All 11 Two-port questions, with notes

GATE 2023 EC Q.46: solution and answer

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